00:02
Okay, so here we want to find the rate law for this reaction and then also find the value of the rate constant, k.
00:10
So to find the rate law, we know that the rate is going to be equal to k times the concentration of n2 and h2, and these will be raised to certain exponents based on the order of these reactants to the total reaction.
00:39
So to find these exponents, we'll need to use the data that we have here in the table, and then from that, once we find those exponents, then we can calculate k.
00:51
So to find these exponents, we need to use the data we have in this table to eliminate one of our variables, because we can't solve with two unknown variables.
01:01
So what we can do is we can find two reactions where one of our reactants is kept constant, and use that so we can cancel it out in our equation.
01:10
So if we pick reactions 1 and 2, the rate of the first one is 9 times 10 to the 4th.
01:22
This is equal to k times 1 .67 raised to the x and 1 .22 raised to the y.
01:31
We divide this by the data for experiment 2, so 1 .48 times 10 to the 5th and 2 .01.
02:04
So now we can go ahead and cancel out what's common here.
02:07
So k and our x term will cancel out and we're left with just the y term.
02:12
So here we have 9 times 10 to the 4th divided by 1 .48 times 10 to the 5th, which comes out to 0 .612.
02:35
And then this is going to be equal to 1 .22 divided by 2 .01 raised to the y.
02:49
So 6 .0, so here if we round our answer we get 0 .612 is equal to 0 .61 raised to the y.
03:24
And so here we know that y is going to be equal to 1 because things raised to the exponent of 1 stay constant.
03:37
So then to find the exponent for x, we want to find two reactions where the concentration of hydrogen is kept constant.
03:45
So we're going to look at experiments 1 and 3.
03:49
And we'll do the same thing.
03:50
So we have 9 .00 times 10 to the fourth, and then we'll plug in the data for experiment three.
04:21
Then we can go ahead and cancel out our k values and our y values, and again, we simplify so that we can have just the x exponent.
04:37
So 9 times 10 to the fourth divided by 2 .64 times 10 to the fifth, and then here we'll simplify on the right side, 1 .67 divided by 2 .86 raised to the x.
05:28
And so here we want to see which one of these is going to, what exponent is going to yield this answer here.
05:39
And so commonly we can plug in 0, 1, and 2 because reactions are usually 0, first, or second order.
05:45
If we plug in 0, then this term turns into 1.
05:53
If we plug in 1 as our exponent, then the number will stay the same...